int x;
int komsuSayisi;//adjanceny matrix
int **arkadas;
int t;
int komsu[24][24];
scanf("%d",&t);
**arkadas = (int **)malloc( t*sizeof( int* )); //allocating rows
for(i=0; i<t; i++)
{
x=0;
arkadas[i] = (int *)malloc( t*sizeof(int) ); //allocating cow temporarily
for(j=0; j<t; j++)
{
komsu[i][j]=fark(kelime[i],kelime[j]); //fark returns 1 or 0.
//so i put those 1 ones to another matrix,arkadas
if(komsu[i][j]==1){
komsuSayisi++;
arkadas[i][x]=j;
x++;
}
arkadas[i] = (int *) realloc(arkadas[i], x);
//real allocating here
}
It gives error and shut downs.There is nothing wrong. What i want is adjanceny is too big to search so i will easily search the "1" ones with this matrix.
**arkadas = (int **)malloc( t*sizeof( int* ));
should be
arkadas = malloc( t*sizeof( int* ));
**arkadas dereferences an uninitialised pointer, resulting in you trying to write to an unpredictable address. You don't own the memory at this address so it isn't safe to try and write to it.
The second form assigns the address of an array of pointers to the local variable arkadas; this is what need to do.
Later in your program
if(komsu[i][j]==1){
komsuSayisi++;
arkadas[i][x]=j;
x++;
}
arkadas[i] = (int *) realloc(arkadas[i], x);
code inside the if condition attempts to write to arkadas[i] before you allocate it. This also invokes undefined behaviour and will likely crash. You can avoid the crash by removing the line arkadas[i][x]=j; and swapping your realloc call for malloc (you need the address of a previous allocation before you can call realloc)
if(komsu[i][j]==1){
komsuSayisi++;
x++;
}
arkadas[i] = malloc(x*sizeof(int));
I see
komsuSayisi++;
You didn't paste the whole code but probably thats what is crashing your program... I don't see any initializing previous to that increment....
this plus the deferentiation posted on the other answer
Related
I am debugging some code written by someone else which has several memory leaks detected by Valgrind. Part of the code which I believe is causing the problems involves void pointers:
int noElements = 100;
void** ptr = malloc( noElements * sizeof(void*));
for(int i = 0; i < noElements; i++)
{
ptr[i] = malloc(sizeof(void*));
}
//Operation filling the array and doing calculation
free(ptr); //This will not clear the underlying array causing the memory leak?
I created a small test program to check the error is where I thought it was (not freeing each ptr[i] before calling free(ptr);), but I am getting errors when trying to free elements. My test program looks like this:
int main()
{
int num = 10000000; //Large array so I can clearly see memory usage
void** ptr = (void**) malloc(sizeof(void*) * num);
for (int i = 0; i < num; i++)
{
ptr[i] = malloc(sizeof(void*));
ptr[i] = &i; //If I never point elements at anything, this code works
}
for (int i = 0; i < num; i++)
{
free(ptr[i]); //Crashes here if the elements have been pointed at a variable
}
free(ptr);
return 0;
}
Why would calling free on each array element cause the program to crash if the pointers have actually been assigned?
In your code
ptr[i] = &i;
creates three issues.
It makes the actual returned pointer by malloc() to be lost (you're overwriting it), so you have no shot at free()-ing it later, leading to memory leak.
i is a local scoped variable (the scope of for loop body), and you're storing the address of the variable to be used outside the scope (i.e., after the lifetime is over). Outside the scope attempt to access the address of it will invoke undefined behavior.
(Ignoring point 2) Attempt to free() the ptr[i] will cause undefined behavior again, as the pointer is not returned by memory allocator function.
Your pointers have been assigned a value which is not what you want.
for (int i = 0; i < num; i++)
{
ptr[i] = malloc(sizeof(void*));
ptr[i] = &i; //If I never point elements at anything, this code works
}
First you assign a valid memory address that is returned by malloc().
This address could be free'd using free().
But then you assign another address. The address of your local variable i. This memory location was not allocated using malloc() and hence cannot be free'd. By assigning this address you've lost the dynamically allocated address causing a memory leak.
Your comment is wrong. You already assigned a value that ptr[i] points to some valid address. You don't need to assign again. And you should not.
I wrote a C code that usea a matrix of double:
double y[LENGTH][4];
whith LENGTH=200000 I have no problem.
I have to increase the numbers of rows to LENGTH=1000000 but when I enter this value and execute the program it returns me segmentation fault.
I tried to allocate more memory using malloc:
double **y = (double **)malloc(LENGTH * sizeof(double*));
for(int i = 0; i < LENGTH; i++){
y[i] = (double *)malloc(4 * sizeof(double));
}
I run the the code above and after some second of calculations it still gives me "segmentation fault".
Could anyone help me?
If you want a dynamic allocated 2D array of the specified row-width, just do this:
double (*y)[4] = malloc(LENGTH * sizeof(*y));
There is no need to malloc each row in the matrix. A single malloc and free will suffice. Only if you need dynamic row width (each row can vary in width independent of others) or the column count is arbitrary should a nested malloc loop be considered. Neither appears to be your case here.
Notes:
Don't cast malloc in C programs
Be sure to free(y); when finished with this little tryst.
The reason your statically allocated array is segfaulting with a million elements is (presumably) because it's being allocated on the stack. To have your program have a larger stack, pass appropriate switches to your compiler.
ProTip: You will experience less memory fragmentation and better performance if you flip your loop around, allocating
(double *)malloc(LENGTH * sizeof(double));
four times. This will require changing the order of your indices.
I ran the the code with this definition and after some second of calculations it still gives me "segmentatin fault"
If you're getting a segmentation fault after allocating the memory, you're writing outside of your memory bounds.
I run this code
#include <stdio.h>
#include <stdlib.h>
// We return the pointer
int **get(int N, int M) /* Allocate the array */
{
/* Check if allocation succeeded. (check for NULL pointer) */
int i, **table;
table = malloc(N*sizeof(int *));
for(i = 0 ; i < N ; i++)
table[i] = malloc( M*sizeof(int) );
return table;
}
void free2Darray(int** p, int N) {
int i;
for(i = 0 ; i < N ; i++)
free(p[i]);
free(p);
}
int main(void)
{
const int LENGTH = 1000000;
int **p;
p = get(LENGTH, 4);
printf("ok\n");
free2Darray(p ,LENGTH);
printf("exiting ok\n");
return 0;
}
and it was executed normally.
I got the code from my pseudo-site.
You should not cast what malloc returns. Why?
Also notice, that since you need a dynamic allocation only for the number of rows, since you know the number of columns. So, you can modify the code yourself (so that you have some fun too. :) )
I hope you didn't forget to **free** your memory.
I am trying to make a generic function in C that takes a 2D array of ANY type and copies it into a contiguous memory block. ( I need this function for Aggregate operations on MPI on my complex datatypes).
Imagine I have the following integer array
int n = 5;
int m = 6;
int** int_array = (int**) malloc(n* sizeof(int*));
for (int i = 0; i < n; i++ )
int_array[i] = (int *) malloc(m * sizeof(int) );
In this type of memory allocation one cannot, in principle, hope to access the , say i,j-th entry of int_array using the following pointer arithmetics
int value = (*lcc)[i*m+j];
Therefore I implemented a function that basically allocates a new memory block and neatly orders the entries of int_array so that the above indexing should work.
void linearize(char*** array, int n, int m,unsigned int size_bytes){
char* newarray = (char*)malloc(m*n*size_bytes);
//copy array!
for (int i = 0;i<n;i++)
for(int j = 0;j<m*size_bytes;j++)
{
newarray[i*m*size_bytes+j] = (*array)[i][j];
}
//swap pointers and free old memory!
for (int i = 0;i<n;i++)
{
char * temp = (*array)[i];
(*array)[i] = newarray + i*m*size_bytes ;
free(temp);
}
}
I wanted to make the above function to work with any kind of array type, hence I used char pointers to do operations byte by byte. I tested the function and so far it works, but I am not sure about memory deallocation.
Does free(temp) free the whole memory pointed to by int_array[i], that is the m*sizeof(int) bytes accessible from int_array[i] or only the first m bytes (since it thinks that our array is of type char rather than in) ? Or simply put, "Does the linearize function induce any memory leaks? "
Thank you in advance!
*EDIT*
As suggested by Nicolas Barbey, I ran a valgrind checks for memory leaks and it found none.
So to summarize the main points that I found difficult to understand about the behaviour of the program were:
in the function linearize does the following code induce memory leaks:
char * temp = (*array)[i];
(*array)[i] = newarray + i*m*size_bytes ;
free(temp);
NO!! somehow gnu compiler is smart enough to know how many bytes pointed to by "temp" to free. Originally I was afraid that if I array[i] is a pointer of type int , for example, that points to a memory location with say 5 ints = 5*4 bytes, the free(temp) is going to free only the first five bytes of that memory.
Another point to make is : how to free the already linearized array? that is if you have:
// first initialize the array.
int** array = (int**)malloc(5*sizeof(int*);
for(int i = 0; i< 5;i++)
array[i] = ( int* ) malloc(5*sizeof(int));
//now a call to linearize
linearize(&array,5,5,sizeof(int));
... do some work with array ....
// now time to free array
free(array[0]);
free(array);
//suffices to free all memory pointed to by array[i] and as well as the memory allocated
// for the pointers.
Thanks for the discussion and the suggestions.
You need to call free() exactly one call per malloc() inorder to be no memory leaks. Which means in your case int_array is passed to linearize function allocates a block of memory other than int_array allocation, therefore you need to loop over int_array[i] freeing each int* that you traverse followed by free'ing int_array itself. Also you need to free block created in linearize function too.
Here is a slightly slimmer version using actual two dimensional arrays:
void * linearize(void** array, int n, int m,unsigned int size_bytes){
char (*newarray)[m * size_bytes] = malloc(m*n*size_bytes);
//copy array!
int i;
for (i = 0;i<n;i++) {
memcpy(newarray[i], array[i], sizeof(*newarray));
free(array[i]);
}
free(array);
return newarray;
}
Use:
int (*newarray)[m] = linearize(array, n, m, sizeof(**int_array));
int value = newarray[i][j];
// or
value = newarray[0][i*m + j];
// or
value = ((int *)newarray)[i*m + j];
I want to create an integer pointer p, allocate memory for a 10-element array, and then fill each element with the value of 5. Here's my code:
//Allocate memory for a 10-element integer array.
int array[10];
int *p = (int *)malloc( sizeof(array) );
//Fill each element with the value of 5.
int i = 0;
printf("Size of array: %d\n", sizeof(array));
while (i < sizeof(array)){
*p = 5;
printf("Current value of array: %p\n", *p);
*p += sizeof(int);
i += sizeof(int);
}
I've added some print statements around this code, but I'm not sure if it's actually filling each element with the value of 5.
So, is my code working correctly? Thanks for your time.
First:
*p += sizeof(int);
This takes the contents of what p points to and adds the size of an integer to it. That doesn't make much sense. What you probably want is just:
p++;
This makes p point to the next object.
But the problem is that p contains your only copy of the pointer to the first object. So if you change its value, you won't be able to access the memory anymore because you won't have a pointer to it. (So you should save a copy of the original value returned from malloc somewhere. If nothing else, you'll eventually need it to pass to free.)
while (i < sizeof(array)){
This doesn't make sense. You don't want to loop a number of times equal to the number of bytes the array occupies.
Lastly, you don't need the array for anything. Just remove it and use:
int *p = malloc(10 * sizeof(int));
For C, don't cast the return value of malloc. It's not needed and can mask other problems such as failing to include the correct headers. For the while loop, just keep track of the number of elements in a separate variable.
Here's a more idiomatic way of doing things:
/* Just allocate the array into your pointer */
int arraySize = 10;
int *p = malloc(sizeof(int) * arraySize);
printf("Size of array: %d\n", arraySize);
/* Use a for loop to iterate over the array */
int i;
for (i = 0; i < arraySize; ++i)
{
p[i] = 5;
printf("Value of index %d in the array: %d\n", i, p[i]);
}
Note that you need to keep track of your array size separately, either in a variable (as I have done) or a macro (#define statement) or just with the integer literal. Using the integer literal is error-prone, however, because if you need to change the array size later, you need to change more lines of code.
sizeof of an array returns the number of bytes the array occupies, in bytes.
int *p = (int *)malloc( sizeof(array) );
If you call malloc, you must #include <stdlib.h>. Also, the cast is unnecessary and can introduce dangerous bugs, especially when paired with the missing malloc definition.
If you increment a pointer by one, you reach the next element of the pointer's type. Therefore, you should write the bottom part as:
for (int i = 0;i < sizeof(array) / sizeof(array[0]);i++){
*p = 5;
p++;
}
*p += sizeof(int);
should be
p += 1;
since the pointer is of type int *
also the array size should be calculated like this:
sizeof (array) / sizeof (array[0]);
and indeed, the array is not needed for your code.
Nope it isn't. The following code will however. You should read up on pointer arithmetic. p + 1 is the next integer (this is one of the reasons why pointers have types). Also remember if you change the value of p it will no longer point to the beginning of your memory.
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#define LEN 10
int main(void)
{
/* Allocate memory for a 10-element integer array. */
int array[LEN];
int i;
int *p;
int *tmp;
p = malloc(sizeof(array));
assert(p != NULL);
/* Fill each element with the value of 5. */
printf("Size of array: %d bytes\n", (int)sizeof(array));
for(i = 0, tmp = p; i < LEN; tmp++, i++) *tmp = 5;
for(i = 0, tmp = p; i < LEN; i++) printf("%d\n", tmp[i]);
free(p);
return EXIT_SUCCESS;
}
//Allocate memory for a 10-element integer array.
int array[10];
int *p = (int *)malloc( sizeof(array) );
At this point you have allocated twice as much memory -- space for ten integers in the array allocated on the stack, and space for ten integers allocated on the heap. In a "real" program that needed to allocate space for ten integers and stack allocation wasn't the right thing to do, the allocation would be done like this:
int *p = malloc(10 * sizeof(int));
Note that there is no need to cast the return value from malloc(3). I expect you forgot to include the <stdlib> header, which would have properly prototyped the function, and given you the correct output. (Without the prototype in the header, the C compiler assumes the function would return an int, and the cast makes it treat it as a pointer instead. The cast hasn't been necessary for twenty years.)
Furthermore, be vary wary of learning the habit sizeof(array). This will work in code where the array is allocated in the same block as the sizeof() keyword, but it will fail when used like this:
int foo(char bar[]) {
int length = sizeof(bar); /* BUG */
}
It'll look correct, but sizeof() will in fact see an char * instead of the full array. C's new Variable Length Array support is keen, but not to be mistaken with the arrays that know their size available in many other langauges.
//Fill each element with the value of 5.
int i = 0;
printf("Size of array: %d\n", sizeof(array));
while (i < sizeof(array)){
*p = 5;
*p += sizeof(int);
Aha! Someone else who has the same trouble with C pointers that I did! I presume you used to write mostly assembly code and had to increment your pointers yourself? :) The compiler knows the type of objects that p points to (int *p), so it'll properly move the pointer by the correct number of bytes if you just write p++. If you swap your code to using long or long long or float or double or long double or struct very_long_integers, the compiler will always do the right thing with p++.
i += sizeof(int);
}
While that's not wrong, it would certainly be more idiomatic to re-write the last loop a little:
for (i=0; i<array_length; i++)
p[i] = 5;
Of course, you'll have to store the array length into a variable or #define it, but it's easier to do this than rely on a sometimes-finicky calculation of the array length.
Update
After reading the other (excellent) answers, I realize I forgot to mention that since p is your only reference to the array, it'd be best to not update p without storing a copy of its value somewhere. My little 'idiomatic' rewrite side-steps the issue but doesn't point out why using subscription is more idiomatic than incrementing the pointer -- and this is one reason why the subscription is preferred. I also prefer the subscription because it is often far easier to reason about code where the base of an array doesn't change. (It Depends.)
//allocate an array of 10 elements on the stack
int array[10];
//allocate an array of 10 elements on the heap. p points at them
int *p = (int *)malloc( sizeof(array) );
// i equals 0
int i = 0;
//while i is less than 40
while (i < sizeof(array)){
//the first element of the dynamic array is five
*p = 5;
// the first element of the dynamic array is nine!
*p += sizeof(int);
// incrememnt i by 4
i += sizeof(int);
}
This sets the first element of the array to nine, 10 times. It looks like you want something more like:
//when you get something from malloc,
// make sure it's type is "____ * const" so
// you don't accidentally lose it
int * const p = (int *)malloc( 10*sizeof(int) );
for (int i=0; i<10; ++i)
p[i] = 5;
A ___ * const prevents you from changing p, so that it will always point to the data that was allocated. This means free(p); will always work. If you change p, you can't release the memory, and you get a memory leak.
I'm having some issues with producing an int matrix without creating memory leaks. I want to be able to make a given (global) matrix into any size dynamically via read_matrix(). But then i want to be able to free the memory later on. So in my main method the second printf should result in a bus error since it should not have any memory allocated to it. How would i go about creating this?
int** first_matrix;
int** second_matrix;
int** result_matrix;
int** read_matrix(int size_x, int size_y)
{
int** matrix;
matrix = calloc(size_x, sizeof(int*));
for(int i = 0;i<size_x;i++) {
matrix[i] = calloc(size_y, sizeof(int));
}
for(int i = 0;i<size_x;i++) {
for(int j = 0;j<size_y;j++) {
matrix[i][j] = i*10+j;
}
}
return matrix;
}
int main(int stackc, char** stack)
{
first_matrix = read_matrix(10,10);
printf("9:3 %d - 4:6 %d \n", first_matrix[9][3], first_matrix[4][6]);
free(*first_matrix);
free(first_matrix);
printf("9:3 %d - 4:6 %d \n", first_matrix[9][3], first_matrix[4][6]);
}
Just because the memory has been free'd doesn't mean you can't access it! Of course, it's a very bad idea to access it after it's been free'd, but that's why it works in your example.
Note that free( *first_matrix ) only free's first_matrix[0], not the other arrays. You probably want some kind of marker to signify the last array (unless you will always know when you free the outer array how many inner arrays you allocated). Something like:
int** read_matrix(int size_x, int size_y)
{
int** matrix;
matrix = calloc(size_x, 1+sizeof(int*)); // alloc one extra ptr
for(int i = 0;i<size_x;i++) {
matrix[i] = calloc(size_y, sizeof(int));
}
matrix[size_x] = NULL; // set the extra ptr to NULL
for(int i = 0;i<size_x;i++) {
for(int j = 0;j<size_y;j++) {
matrix[i][j] = i*10+j;
}
}
return matrix;
}
Then when you're freeing them:
// keep looping until you find the NULL one
for( int i=0; first_matrix[i] != NULL; i++ ) {
free( first_matrix[i] );
}
free( first_matrix );
You need to free each row individually:
void free_matrix(int **matrix, int size_x)
{
for(int i = 0; i < size_x; i++)
free(matrix[i]);
free(matrix);
}
Freeing the memory doesn't make it go away, it just means that another allocation might grab that same chunk of memory. Whatever you put in it will still be there until something else overwrites it.
Also, you're not freeing everything you allocated. You're only freeing the array of pointers and the first row. But even if you free everything correctly, you would still have the same effect.
If you want to create a "bus error" you need to point to memory that doesn't belong to your process. Why do you want to do that anyway?
You only freed the first row (or column) of first_matrix. Write another function like this:
void free_matrix(int **matrix, int rows)
{
int i;
for(i=0; i<rows; i++)
{
free(matrix[i]);
}
free(matrix);
}
You might want to make the matrix into a struct to store it's row and column count.
I recommend using valgrind to track down unfree'd memory, as opposed to trying to make a bus error occur. It rocks for lots of other stuff as well.
Sam
You're getting memory leaks because you're freeing the first row of the matrix and the list of rows, but none of the 1 to nth rows. You need to call free in a loop.
There are a couple of alternatives, however:
- Allocate sizeof(int*)rows + rowscols*sizeof(int) bytes and use the first bytes for the row pointers. That way, you only have a single chunk of memory to free (and it's easier on the allocator, too)
- Use a struct that contains the number of rows. Then you can avoid the row list altogether (saving memory). The only downside is that you have to use a function, a macro, or some messy notation to address the matrix.
If you go with the second option, you can use a struct like this in any C99 compiler, and again only have to allocate a single block of memory (of size numints*sizeof(int)+sizeof(int)):
struct matrix {
int rows;
int data[0];
}
The concept you are missing here, is that for every calloc, there must be a free.
and that free must be applied to the pointer passed back from calloc.
I recommend you create a function (named delete_matrix)
that uses a loop to free all of the pointers that you allocate in here
for(int i = 0;i < size_x;i++) {
matrix[i] = calloc(size_y, sizeof(int));
}
then, once that is done, free the pointer allocated by this.
matrix = calloc(size_x, sizeof(int*));
The way you are doing it now,
free(*first_matrix);
free(first_matrix);
won't do what you want it to do.